The present invention relates to digital communication and more particularly to protocols for communicating data originating from sources having disparate data rates over a shared medium.
Trends in digital communication point toward a common transmission medium providing both high data rate services such as digital video and low data rate services such as voice. Internet access is inherently a mixed service. Upstream requests for information typically include minimal data while downstream traffic may include graphics or even live video.
Specific examples of such a common transmission medium include a wireless local loop (WLL) that substitutes for the local telephone loop and provides additional high data rate services such as video and Internet access. Another example is a CATV network that has been updated to provide high data rate services and voice service.
A key objective is maximizing efficiency in use of bandwidth. The available bandwidth is shared among multiple data communication devices. When a data communication device is allocated all or part of the available bandwidth, it should make efficient use of its allocation. Depending on the protocols and modulation systems used, a certain percentage of the data is devoted to network operation rather than customer service. This is referred to as overhead. Consider a network where packets of information are communicated in successive frames and:
dp=Payload data (bits)—the number of payload bits contained in a frame,
r=Data rate (bits/sec), proportional to spectrum used (Hz). Data rate refers to the rate at which information is communicated through the wireless medium. Information rate roughly represents the rate of generation of payload data.
tf=Frame time (sec)—the duration of the smallest unit of time that may be allocated to a data communication device for transmission on the shared medium. Note that a packet of like data, such as voice or data, may be transmitted in a single frame or may be divided among many frames.
tg=Overhead time (sec), including guard time, training, and synchronization that is required for each frame.
The system efficiency associated with the overhead given by tg is
  eff  =                              d          p                          r          ⁡                      (                                          t                g                            +                              t                f                                      )                              ·      This        ⁢                  ⁢    value    ⁢                  ⁢    of    ⁢                  ⁢    efficiency    ⁢                  ⁢    reaches    ⁢                  ⁢    100    ⁢                  ⁢    %    ⁢                              ⁢                            ⁢    when    ⁢                  ⁢    the    ⁢                  ⁢    overhead  time is zero and the frame time equals the payload data divided by the data rate (when the frame time is exactly the time required to transmit the payload data at the transmission rate).
The network designer is left free to vary frame time to maximize efficiency. However, it is difficult to reconcile the needs of different traffic types. Consider choosing one frame time to accommodate both low information rate voice traffic and high information rate data traffic. Due to transmission latency requirements, voice traffic requires frequent frame transmissions to reduce latency. Hence, voice requires a short frame time. Furthermore, the amount of data to be sent in these frames is small since voice is low information rate. If long frames are sent, voice traffic is insufficient to fill each frame, resulting in wasted bandwidth. However, sending small frames incurs a different type of bandwidth loss. The fixed overhead associated with each frame substantially reduces spectral efficiency. This becomes particularly significant for high information rate traffic, where data must be divided over many frames instead of being efficiently transmitted in long frames. This conflict can be described in an example.
Consider choosing a frame time to efficiently transmit both a 64-byte voice packet and a 1000-byte data packet. Assume an overhead time of 3 us (tg=3 us), a data rate of 30 Mbits/sec, and two candidate frame times of 17 us and 267 us. For a frame time of 17 us, the efficiency for a 64-byte packet of data is,
      eff    64    =                    d        p                    r        ⁡                  (                                    t              g                        +                          t              f                                )                      =                  512                  30          ×                      10            6                    ⁢                      (                                          (                                  3                  +                  17                                )                            ×                              10                                  -                  6                                                      )                              =                        17                      3            +            17                          =                  85          ⁢                                          ⁢                      %            .                              
Since the frame time is exactly the amount of time required to transmit 64 bytes at a 30 Mbit/sec rate, this is the maximum efficiency at this data rate. A 1000-byte packet would be spread among 64-byte transmission opportunities corresponding to individual frames. Hence the efficiency for a 1000 byte packet is approximately the same as for the 64-byte packet. (To be precise, the efficiency is slightly less than 85% since the final frame is not fully utilized.)
Increasing the frame time can increase this efficiency by reducing the overhead. For example, a frame time of 267 us results in close to 99% efficiency for a 1000 byte packet,
      eff    1000    =                    d        p                    r        ⁡                  (                                    t              g                        +                          t              f                                )                      =                  8000                  30          ×                      10            6                    ⁢                      (                                          (                                  3                  +                  267                                )                            ×                              10                                  -                  6                                                      )                              =                        267                      3            +            267                          =                  99          ⁢                                          ⁢                      %            .                              
Unfortunately, this large frame time causes severe inefficiency for the 64-byte packet because a large portion of the frame is left unutilized. Here it is assumed that because of latency requirements, it is not feasible to collect multiple 64-byte packets to fill a frame. For example, it would take a 64 kbps voice source over 125 ms to fill a 1000-byte frame, which results in intolerable latency. Allowing 8 ms of latency for the collection of one 64-byte packet, the long frame capable of support6ing 1000 bytes which carries 64-byte packet has very poor efficiency:
      eff    64    =                    d        p                    r        ⁡                  (                                    t              g                        +                          t              f                                )                      =                  512                  30          ×                      10            6                    ⁢                      (                                          (                                  3                  +                  267                                )                            ×                              10                                  -                  6                                                      )                              =                        17                      3            +            267                          =                  6          ⁢                                          ⁢                      %            .                              
No single choice of frame time leads to efficient use of the spectrum for both high data rate and low data rate traffic.